How do you find the definite integral for: [(6x^2+2) / sqrt(x)] dx for the intervals [1, 5]?

2 Answers
May 23, 2017

~~136.71

Explanation:

"rewrite as"

(6x^2)/(x^(1/2))+2/x^(1/2)=6x^(3/2)+2x^(-1/2)

rArrint_1^5(6x^(3/2)+2x^(-1/2))dx

"integrate each term using the "color(blue)"power rule"

• int(ax^n)=a/(n+1)x^(n+1); n!=-1

=[12/5x^(5/2)+4x^(1/2)]_1^5

=(12/5. 5^(5/2)+4. 5^(1/2))-(12/5 .1+4.1)

=60sqrt5+4sqrt5-32/5

=64sqrt5-32/5~~136.71" to 2 dec. places"

May 23, 2017

64sqrt5-6.4

Explanation:

Let, I=int_1^5 (6x^2+2)/sqrtx dx.

We know that, intx^ndx=x^(n+1)/(n+1), n!=-1.

:. I=int_1^5(6x^2+2)/sqrtxdx,

=int_1^5(6x^2+2)*x^(-1/2)dx,

=int_1^5{(6x^(2-1/2)+2x^(-1/2)}dx,

=6int_1^5x^(3/2)dx+2int_1^5x^(-1/2)dx,

=6[x^(3/2+1)/(3/2+1)]_1^5+2[x^(-1/2+1)/(-1/2+1)]_1^5,

=6*2/5[x^(5/2)]_1^5+2*2[x^(1/2)]_1^5,

=12/5[5^(5/2)-1^(5/2)]+4[5^(1/2)-1^(1/2)]

=12/5(25sqrt5-1)+4(sqrt5-1),

=60sqrt5-12/5+4sqrt5-4,

rArr I=64sqrt5-6.4

Enjoy Maths.!